91Ó°ÊÓ

The residue is given by the relation: $R=\frac{1}{\left(n-1\right)!}\underset{\mathrm{z}→{\mathrm{z}}_0}{\lim }\frac{{\mathrm{d}}^{\mathrm{n}-1}}{{\mathrm{dz}}^{\mathrm{n}-1}}\left(\mathrm{z}-{\mathrm{z}}_0\right)\mathrm{f}\left(\mathrm{z}\right)$

(a) The residue is given by the relation,

$R=\frac{1}{\left(\mathrm{n}-1\right)!}\underset{\mathrm{z}→{\mathrm{z}}_0}{\lim }\frac{{\mathrm{d}}^{\mathrm{n}-1}}{{\mathrm{dz}}^{\mathrm{n}-1}}\left(\mathrm{z}-{\mathrm{z}}_0\right)\mathrm{f}\left(\mathrm{z}\right)$

so, if $\mathrm{f}\left(\mathrm{z}\right)$is given as follow:

$\mathrm{f}\left(\mathrm{z}\right)=\frac{\cos \mathrm{z}}{{\left(2\mathrm{z}-\mathrm{Ï€}\right)}^4}$

The substitution in (1) yields as follows:

$$\begin{gathered} R=\frac{1}{\left(3!\right)\left(2^4\right)}\underset{z→\frac{\mathrm{π}}{2}}{\lim }\frac{{\mathrm{d}}^2}{{\mathrm{dz}}^2}{\left(\mathrm{z}-\frac{\mathrm{z}}{2}\right)}^4\frac{\cos \mathrm{z}}{{\left(\mathrm{z}-{\displaystyle \frac{\mathrm{π}}{2}}\right)}^4} \\ =\frac{1}{\left(3!\right)\left(2^4\right)}\underset{z→\frac{\mathrm{π}}{2}}{\lim }\frac{{\mathrm{d}}^3\cos \left(\mathrm{z}\right)}{{\mathrm{dz}}^3} \\ =\frac{1}{\left(3!\right)\left(2^4\right)}\underset{z→\frac{\mathrm{π}}{2}}{\lim }\frac{{\mathrm{d}}^2\left[-\sin \left(\mathrm{z}\right)\right]}{{\mathrm{dz}}^2} \\ =\frac{1}{\left(3!\right)\left(2^4\right)} \end{gathered}$$

simplify further as follow:

$$\begin{gathered} R=\frac{1}{\left(3!\right)\left(2^4\right)} \\ =\frac{1}{\left(3!\right)\left(2^4\right)}\underset{z→\frac{\mathrm{π}}{2}}{\lim }\frac{\mathrm{d}\left(-\cos \left(\mathrm{z}\right)\right)}{\mathrm{dz}} \\ =\frac{1}{96} \end{gathered}$$

Thus the residue islocalid="1664378049682" $\frac{1}{96}$.

If the function is given as,

$\mathrm{f}\left(\mathrm{z}\right)=\frac{2{\mathrm{z}}^2+3\mathrm{z}}{\mathrm{z}-1}$

If we need to get the residue at infinity, then the residue is given as follows:

$\mathrm{R}\left(\mathrm{z}→∞\right)=\mathrm{R}\left(\frac{\mathrm{f}\left({\displaystyle \frac{1}{\mathrm{z}}}\right)}{{\mathrm{z}}^2},\mathrm{z}→0\right)$

So, let's replace $z→\frac{1}{z}$ and obtain:

$$\begin{gathered} =\frac{2{\left({\displaystyle \frac{1}{z}}\right)}^2+3\left({\displaystyle \frac{1}{z}}\right)}{\left({\displaystyle \frac{1}{z}}\right)-1} \\ =\frac{2+3z}{z\left(1-z\right)} \end{gathered}$$

So, the residue at infinity is given as follows:

$R\left(z→∞\right)=-R\left(\frac{2+3z}{z^3\left(1-z\right)},z→0\right)$

So, the residue is given as follows:

$$\begin{gathered} R\left(z→∞\right)=-\frac{1}{2!}\underset{z→∞}{\lim }\frac{d^2}{d{z}^2}\frac{2+3z}{\left(1-z\right)} \\ =\frac{1}{2!}\underset{z→∞}{\lim }\frac{d}{dz}\left(-\frac{5}{{\left(z-1\right)}^2}\right) \\ =-\frac{5}{2!}\underset{z→∞}{\lim }\left(\frac{-2\left(z-1\right)}{{\left(z-1\right)}^4}\right) \\ =-\frac{5}{2!}×2 \\ =-5 \end{gathered}$$

Thus, the residue is –5.

(c)

If the function is given as follows:

$$\begin{gathered} f\left(z\right)=\frac{z^3}{1+32z^5} \\ f\left(z\right)=\frac{1}{32\left(z^5+{\displaystyle \frac{1}{32}}\right)} \end{gathered}$$

The function has a simple pole at$z=-0.5$ , so the residue is given as follows:

$$\begin{gathered} R(z=-0.5)=\frac{P\left(z_0\right)}{Q\left(z_0\right)} \\ ={\left[\frac{z^3}{\left(32\right)\left(5\right)z^4}\right]}_{z→-1/2} \\ =\frac{1}{\left(32\right)\left(5\right)\left(-{\displaystyle \frac{1}{2}}\right)} \\ =-\frac{1}{80} \end{gathered}$$

Thus, the residue is –1/80.

If the function is given by,

$f\left(z\right)=csc\left(2z-3\right)$

First, let us state the series of $csc\left(z\right)$, which is given as follows:

localid="1664377936360" $csc\left(z\right)=\frac{1}{z}+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z}{15120}+....$$csc\left(z\right)=\frac{1}{z}+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+........$

Let's replace $z→2z-3$ which is given as follows:

$csc\left(2z-3\right)=\frac{1}{2z-3}+\frac{2z-3}{6}+\frac{7{\left(2z-3\right)}^3}{360}+....$

So it can also be written in the form:

$csc\left(2z-3\right)=\frac{1}{2\left(z-{\displaystyle \frac{3}{2}}\right)}+\frac{2\left(z-{\displaystyle \frac{3}{2}}\right)}{6}+....$

The residue at $z_0=3/2$ is the factor of which is .